Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 11, Problem 5P
If you examine Fig. 11.5 closely, you will note that in some regions, such as between
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What would be a medical advantage of having a catalog of SNPs?
You learned in Problem 21 in Chapter 7 that theneurodegenerative disease ALS can be caused by expansion of a hexanucleotide repeat region (5′-GGGGCC-3′)outside of the open reading frame (but within the firstintron) of the gene called C9ORF72. While a normalC9ORF72 allele has 2–23 copies of the hexanucleotiderepeat unit, dominant disease-causing alleles have hundreds or even thousands of copies. Researchers observed that the first intron of theC9ORF72 disease allele is transcribed not only fromthe normal template strand of DNA, but also from thenontemplate strand. Even more unusual, both types ofrepeat-region transcripts are translated in all six readingframes in an AUG-independent manner—a processcalled repeat-associated non-ATG translation, or RANtranslation. These discoveries led to the hypothesisthat the proteins made from the repeats mightcontribute to ALS.a. What polypeptides are made from the repeat-regiontranscripts?b. According to the RAN translation hypothesis, whyare…
Let’s suppose you make a transposon library of the cellulose-secreting bacterium Komagataeibacter xylinus, with the goal of finding mutants that produce higher than normal amounts of cellulose, which would be useful industrially. However, despite your best efforts you are unable to isolate any transposon mutants that make more cellulose than the wild-type strain.Why might this have failed? List as many reasons as you can think of.
Chapter 11 Solutions
Genetics: From Genes to Genomes
Ch. 11 - Choose the phrase from the right column that best...Ch. 11 - Would you characterize the pattern of inheritance...Ch. 11 - Would you be more likely to find single nucleotide...Ch. 11 - A recent estimate of the rate of base...Ch. 11 - If you examine Fig. 11.5 closely, you will note...Ch. 11 - Approximately 50 million SNPs have thus far been...Ch. 11 - Mutations at simple sequence repeat SSR loci occur...Ch. 11 - Humans and gorillas last shared a common ancestor...Ch. 11 - In 2015, an international team of scientists...Ch. 11 - Using PCR, you want to amplify an approximately 1...
Ch. 11 - Prob. 11PCh. 11 - The previous problem raises several interesting...Ch. 11 - You want to make a recombinant DNA in which a PCR...Ch. 11 - You sequence a PCR product amplified from a...Ch. 11 - Prob. 15PCh. 11 - The trinucleotide repeat region of the Huntington...Ch. 11 - Sperm samples were taken from two men just...Ch. 11 - Prob. 18PCh. 11 - a. It is possible to perform DNA fingerprinting...Ch. 11 - On July 17, 1918, Tsar Nicholas II; his wife the...Ch. 11 - The figure that follows shows DNA fingerprint...Ch. 11 - Microarrays were used to determine the genotypes...Ch. 11 - A partial sequence of the wild-type HbA allele is...Ch. 11 - a. In Fig. 11.17b, PCR is performed to amplify...Ch. 11 - The following figure shows a partial microarray...Ch. 11 - Scientists were surprised to discover recently...Ch. 11 - The microarray shown in Problem 25 analyzes...Ch. 11 - The figure that follows shows the pedigree of a...Ch. 11 - One of the difficulties faced by human geneticists...Ch. 11 - Now consider a mating between consanguineous...Ch. 11 - The pedigree shown in Fig. 11.22 was crucial to...Ch. 11 - You have identified a SNP marker that in one large...Ch. 11 - The pedigrees indicated here were obtained with...Ch. 11 - Approximately 3 of the population carries a mutant...Ch. 11 - The drug ivacaftor has recently been developed to...Ch. 11 - In the high-throughput DNA sequencing protocol...Ch. 11 - A researcher sequences the whole exome of a...Ch. 11 - As explained in the text, the cause of many...Ch. 11 - Figure 11.26 portrayed the analysis of Miller...Ch. 11 - A research paper published in the summer of 2012...Ch. 11 - Table 11.2 and Fig. 11.27 together portray the...Ch. 11 - The human RefSeq of the entire first exon of a...Ch. 11 - Mutations in the HPRT1 gene in humans result in at...Ch. 11 - Prob. 44P
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- In addition to the standard base-paired helical structures, DNA can form X-shaped hairpin structures called cruciforms in which most bases are involved in Watson–Crick pairs. Such structures tend to occur at sequences with inverted repeats. Draw the cruciform structure formed by the DNA sequence TCAAGTCCACGGTGGACTTGC.arrow_forwardThe D1S80 locus is located on human chromosome 1 and is characterized by a repeating 16 base pair (bp) sequence. Alleles for this locus vary depending on the number of repeats present, thus affecting the size of the locus. The D1S80 locus also contains two conserved sequences, a 32bp sequence at one end and a 113bp sequence at the other end. If the DNA of an individual is targeted for D1S80 amplification, and one of the resulting amplicons is approximately 785bp in size, how many repeats would be present in this D1S80 allele? The amplicon of interest is indicated by a red arrow in the diagram below.arrow_forwardList some potential biological effects of SNPs.arrow_forward
- The human RefSeq of the entire first exon of a geneinvolved in Brugada syndrome (a cardiac disordercharacterized by an abnormal electrocardiogram andan increased risk of sudden heart failure) is:5′ CAACGCTTAGGATGTGCGGAGCCT 3′The genomic DNA of four people (1–4), three ofwhom have the disorder, was subjected to singlemolecule sequencing. The following sequences represent all those obtained from each person. Nucleotidesdifferent from the RefSeq are underlined. Individual 1:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGCGGAGACT 3′Individual 2:5′ CAACGCTTAGGATGTGAGGAGCCT 3′Individual 3:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGGCGGAGCCT 3′Individual 4:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGTGGAGCCT 3′a. The first exon of the RefSeq copy of this gene includes the start codon. Write as much of the aminoacid sequence of the encoded protein as possible,indicating the N-to-C polarity.b. Are any of these individuals homozygotes? If so,which person and what allele?c. Is…arrow_forward24. 23andme is a direct to consumer genotyping company that uses a microarray to genotype people on 1 million known SNPS in the genome. A man who was adopted wants to know about his ancestry, and purchases a 23andme kit for himself. He finds that of each autosome pair, exactly one is classified as 100% Sub-Saharan African, while the other is a mixture of various East Asian and European components. He also finds that his X chromosome is 100% Sub-Saharan African. What can this man confidently infer about his recent ancestry? a. His mother was of Sub-Saharan African descent b. His father was of Sub-Saharan African descent c. He has no Neanderthal ancestry d. He has no mutations relative to his parents e. 23andme mixed up two samples and he can't infer anythingarrow_forwardThe human genome contains thousands of sequences known as small open reading frames, some of which encode proteins of about 30 amino acids. What is the minimum number of nucleotides required to encode such a protein?arrow_forward
- In DNA-hybridization experiments on six species of plants in the genus Vicia, DNA was isolated from each of the six species, denatured by heating, and sheared into small fragments (W. Y. Chooi. 1971. Genetics 68:213–230). In one experiment, DNA from each species and from E. coli was allowed to renature. The graph shows the results of this renaturation experiment. Q. Notice that, for the Vicia species, the rate of renaturation is much faster in the first hour and then slows down. What might cause this initial rapid renaturation and the subsequent slowdown?arrow_forwardA method for detecting methylated CpGs involvesthe use of a chemical called bisulfite, which convertscytosine to uracil but leaves methylated cytosine untouched. You want to know whether a particularCpG dinucleotide at one location in the genome ismethylated on one or both strands in a tissue sample.The genomic sequence containing this CpG is:5’...TCCATCGCTGCA…3’. You take genomicDNA from the sample tissue, treat it exhaustivelywith bisulfite, and then use flanking primers toPCR-amplify the region including this CpGdinucleotide. You then want to Sanger sequence(see Fig. 9.7) the amplified PCR product. a. After you treat genomic DNA with bisulfite, the twoDNA strands will melt into single strands. Why?b. Your answer to part (a) introduces a potential complication, because if you do not account for this result of bisulfite treatment, the PCR primers willnot amplify the DNA. What special considerationswould be necessary when you design your PCRprimers for this experiment? Could one pair…arrow_forwardPeople who carry a theoretical genetic disorder (called B-disease) can be identified from a 2kb DNA sequence.People who carry this genetic disorder have a single nucleotide polymorphism that results in a change of GTATCC to GGATCC, a site that only occurs once at nucleotide number 750 in this DNA sequence. Answer the following questions based on the information provided. (a) How can you develop a simple molecular test to identify the genetic disorder? (b) If you have carried out the molecular test (based on the information above) on a 100 individual and found that 24 were healthy (BB) and 26 were carriers (bb); 1) What is the ratio of heterozygous? 2) Show how can you identify the three types from the agarose gelarrow_forward
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